YY transformer with floating primary

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Mivey,

There is a rather simple way to illustrate the folly of operating an YY-Xfmr with its primary neutral isolated! You seem to have the calculating tools, so do you want to try it?

Phil
 
Mivey, use any tool you have to determine: Phase-currents, Iab, Ibc, Ica; Line-currents, Ia, Ib, Ic; & Neutral-Current, In, for a 208/120Y Volt source feeding a Y-connected load, for each of the following cases:


Case 1) Load balanced: 3-ph, 4-w (source & Load neutrals connected); Load imp's are, Zab = Zbc = Zca = 10 Ohms@ 0 Deg!


Case 2) Ld unbalanced: 3-ph, 4-w (source & Load neutrals connected); Load imp's are; Zab=10 Ohms@ 0 Deg; Zbc=10 Ohms@ 30 Deg; Zca=15 Ohms@ 45 Deg!


Case 3) Load balanced: 3-ph, 3-w (source & load neutrals isolated); Load imp's same as Case1 !


Case 4) Ld unbalanced: 3-ph, 3-w (source & load neutrals isolated); Load imp's same as Case 2 !


Regards, Phil
 
mivey said:
You're welcome.

I did put in a wye connected load to make it more interesting. Will work for what?
Click to expand...

Only phase to phase connected loads. My understanding is that as long as no ground fault exists on the secondary a floating HO will go unnoticed?
 
Mivey,

My intent is not to give you further grief! I wanted to illustrate how in floating-neutral circuits, using just static, linear, elements representing load-imbalance, will have a deleterious effect on resultant voltages!


If the above is obvious, then, you must agree that any calculation involving unbalanced magnetic-flux in a Xmr's core, must exacerbate the problem!


Regards, Phil
 
Phil Corso said:
Mivey,

My intent is not to give you further grief! I wanted to illustrate how in floating-neutral circuits, using just static, linear, elements representing load-imbalance, will have a deleterious effect on resultant voltages!


If the above is obvious, then, you must agree that any calculation involving unbalanced magnetic-flux in a Xmr's core, must exacerbate the problem!


Regards, Phil
Click to expand...

Seems like just a fancy way of saying that an unbalanced load on the secondary side will cause more of a neutral point shift on the primary side.

However, I'm just a hick from NC.

P.S. I had to look up what deleterious meant. I hate doin' that.
 
Bugman,

Reur... "I had to look up what deleterious meant. I hate doin' that."

Thank you for the back-handed complement! :)

We've only known each other a few days and you're already smarter! :D

Can't wait 'til you've learned my JIFFY's Method for Problem Solving! :lol:

Your Profile shows that you are a PE! Don't take this as the start of a PE-ing contest, but how long have you been practicing?

Wishing you a Happier Life, Phil
 
Phil Corso said:
Mivey, use any tool you have to determine: Phase-currents, Iab, Ibc, Ica; Line-currents, Ia, Ib, Ic; & Neutral-Current, In, for a 208/120Y Volt source feeding a Y-connected load, for each of the following cases:
Click to expand...
I'm not sure what you want with phase currents on a wye system but you are free to calculate the values from the line currents given below or if needed I will calculate them. Also, for the following I am using the impedance values for the wye connection as I did not feel like making a delta to wye transformation to get wye impedances; it should suffice to make your point.

Phil Corso said:
Case 1) Load balanced: 3-ph, 4-w (source & Load neutrals connected); Load imp's are, Zab = Zbc = Zca = 10 Ohms@ 0 Deg!
Click to expand...
Ia = 11.99<-0.15
Ib = 11.99<-120.15
Ic = 11.99<119.85
In = 0<0

Phil Corso said:
Case 2) Load unbalanced: 3-ph, 4-w (source & Load neutrals connected); Load imp's are; Zab=10 Ohms@ 0 Deg; Zbc=10 Ohms@ 30 Deg; Zca=15 Ohms@ 45 Deg!
Click to expand...
Ia = 11.99<-0.15
Ib = 12.01<-90.14
Ic = 8.007<164.92
In = 10.82<113.03

Phil Corso said:
Case 3) Load balanced: 3-ph, 3-w (source & load neutrals isolated); Load imp's same as Case1 !
Click to expand...
Ia = 11.99<-0.15
Ib = 11.99<-120.15
Ic = 11.99<119.85
In = 0<0

Phil Corso said:
Case 4) Load unbalanced: 3-ph, 3-w (source & load neutrals isolated); Load imp's same as Case 2 !
Click to expand...
Ia = 12.69<19.52
Ib = 8.608<-104.78
Ic = 10.58<157.31
In = 0<0
 
I just want to say this is an awesome thread! Everything I was looking for an then some. I certainly have a new addition (albeit a great one) to my library of knowledge :D:)


One last question. How does the shared magnetic flux of the core change the values, if in any way?
 
Mivey,

You did an excellent job on the simple parts, i.e., current magnitudes and angles for all four cases. But, you left out two calcs necessary for a full analysis: a) Neutral-Shift or Neutral-Displacement; and b) Load over-voltages!

a) What is the Magnitude and Angle of the neutral-shift or neutral-displacement between the original position for the balanced case, say 'N', and the location of the resultant triangle's centroid (*), say 'O'?

b) What are the resultant over-voltages experienced by the unbalanced loads!

Now? Is anyone still interested on how neutral-shift of an unbalanced load impacts a Xfmr's performance!

Regards, Phil

(*) Bugman, I hope this word doesn?t cause another problem for you! :bye:
 
mbrooke said:
One last question. How does the shared magnetic flux of the core change the values, if in any way?
Click to expand...
There is very small increase in primary current in the phase corresponding to the shorted secondary phase as that primary phase current has to flow through the other two primary phases on that condition during primary neutral floating. So there would be no appreciable change in core flux of the transformer.
 
mivey said:
Here is an example with voltages and currents, note the neutral to ground voltage after the fault:

Primary impedance: 4607 amps 3-phase, 3990 amps 2-phase, 3455 amps 1-phase with X/R = 10
Secondary impedance: 1012 amps 3-phase, 876.3 amps 2-phase, 54 amps 1-phase with X/R = 0.17
30 kVA transformer (three 10 kVA units)
Load = 15 kW, 2 kvar at 480 volts wye connected at transformer terminals

PRE-FAULT:

At the source:

VAN = 7199<0
VBN = 7199<-120
VCN = 7199<120
VAB = 12470<29.99
VBC = 12470<-90.01
VCA = 12470<150.00

IA = 0.7009<171.2
IB = 0.7009<51.18
IC = 0.7009<-68.82

SA = 4987+j773.4
SB = 4987+j773.5
SC = 4987+j773.6
P = 14.96 kW
Q = 2.320 kvar
S = 15.14 kVA
pf = 98.82%


At the load:

Van = 275.7<-0.99
Vbn = 275.7<-121.00
Vcn = 275.7<119.00
Vab = 477.6<29.01
Vbc = 477.6<-90.99
Vca = 477.6<149.00

Ia = 18.11<-8.584
Ib = 18.11<-128.6
Ic = 18.11<111.4

Sa = 4949+j659.9
Sb = 4949+j659.9
Sc = 4949+j659.9
P = 14.85 kW
Q = 1.980 kvar
S = 14.98 kVA
pf = 99.12%



Now introduce a bolted secondary c-n fault: Ic = 54.03<109.2

At the source:

VAN = 12180<-31.2
VBN = 12550<-91.73
VCN = 381.2<-168.9
VAB = 12470<29.99
VBC = 12470<-90.02
VCA = 12470<150.00

VAG = 7198<-0.01
VBG = 7200<-120
VCG = 7199<120
VNG = 7085<117.1

IA = 1.186<140
IB = 1.222<79.45
IC = 2.080<-70.78

SA = 14280+j2215
SB = 15150+j2350
SC = 112.1+j784.8
P = 29.54 kW
Q = 5.349 kvar
S = 30.02 kVA
pf = 98.40%


At the load:

Van = 466.6<-32.18
Vbn = 480.6<-92.71
Vcn = 0<0
Vab = 477.6<29.01
Vbc = 480.6<-92.71
Vca = 466.6<147.8

Ia = 30.64<-39.77
Ib = 31.57<-100.3
Ic = 0<0

Sa = 14170+j1890
Sb = 15040+j2005
Sc = 0+j0
P = 29.21 kW
Q = 3.895 kvar
S = 29.47 kVA
pf = 99.12%
Click to expand...
Hi mivey!
What is the primary current for the unloaded, neutral floating Y-Y transformer with one shorted secondary of the above type? Thanks.
 
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